# X intercepts

1. Jan 3, 2008

### david18

Ive got a function of f(x)=x^3+x and I need to use factors to show that the graph crosses the x axis once only.

I just factorised it to x(x^2+1) which isn't very helpful, and if i divide everything by x and complete the square with x^2+1 i get a negative number and stuff....

any help?

2. Jan 3, 2008

### arildno

Oh yes it is!

So, you have found that f(x)=x(x^2+1)

What property, i.e, which equation must an x-intercept of the graph fulfill?

3. Jan 3, 2008

x(x^2+1)=0

4. Jan 3, 2008

### symbolipoint

Good is that you factored the function. the x^2 + 1 factor has no "real" zeros; but the x factor has one real zero, being 0. (zero).

Let me try again in case this helps.
(x^2 + 1) = 0 for what real values of x? For NONE. We do not usually graph complex zeros in two dimensional cartesian plane (at least for our purposes here).

When is x=0 (using the other factor)? when x=0, already shown.

5. Jan 3, 2008

### david18

hi, thanks for the reply, it all makes sense now :)